知识点
示例:用户信息表user_profile
id | device_id | gender | age | university | gpa | active_days_within_30 |
1 | 2138 | male | 21 | 北京大学 | 3.4 | 7 |
2 | 3214 | male | 复旦大学 | 4.0 | 15 | |
3 | 6543 | female | 20 | 北京大学 | 3.2 | 12 |
4 | 2315 | female | 23 | 浙江大学 | 3.6 | 5 |
5 | 5432 | male | 25 | 山东大学 | 3.8 | 20 |
6 | 2131 | male | 28 | 山东大学 | 3.3 | 15 |
7 | 4321 | female | 26 | 复旦大学 | 3.6 | 9 |
示例:question_practice_detail
id | device_id | question_id | result | date |
1 | 2138 | 111 | wrong | 2021-05-03 |
2 | 3214 | 112 | wrong | 2021-05-09 |
3 | 3214 | 113 | wrong | 2021-06-15 |
4 | 6543 | 111 | right | 2021-08-13 |
5 | 2315 | 115 | right | 2021-08-13 |
6 | 2315 | 116 | right | 2021-08-14 |
7 | 2315 | 117 | wrong | 2021-08-15 |
…… |
示例: question_detail
question_id | difficult_level |
111 | hard |
112 | medium |
113 | easy |
115 | easy |
116 | medium |
117 | easy |
34、现在运营想要了解复旦大学的每个用户在8月份练习的总题目数和回答正确的题目数情况,请取出相应明细数据,对于在8月份没有练习过的用户,答题数结果返回0.
select
a.device_id,
a.university,
count(b.question_id) as question_cnt,
sum(if(b.result = 'right', 1, 0)) as right_question_cnt
from
user_profile a
left join (
select
device_id,
question_id,
result,
date
from
question_practice_detail
where
month(date) = 8
and year(date) = 2021
) b on a.device_id = b.device_id
where
a.university = "复旦大学"
group by
a.device_id
返回以下结果:
device_id | university | question_cnt | right_question_cnt |
3214 | 复旦大学 | 3 | 0 |
4321 | 复旦大学 | 0 | 0 |
35、现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况,请取出相应数据,并按照准确率升序输出。
select
c.difficult_level,
--方法1:
sum(if(b.result = 'right', 1, 0)) / count(b.result) as correct_rate
--方法2:
sum(if(qpd.result='right', 1, 0)) / count(qpd.question_id) as correct_rate
--方法3:
count(if(qpd.result='right', 1, null)) / count(qpd.question_id) as correct_rate
from
user_profile a
join question_practice_detail b on a.device_id = b.device_id
join question_detail c on b.question_id = c.question_id
where
a.university = '浙江大学'
group by
difficult_level
order by
correct_rate asc
返回以下结果:
difficult_level | correct_rate |
easy | 0.5000 |
medium | 1.0000 |